Divide the following complex numbers. $ \dfrac{-12+12i}{-3-3i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-3+3i}$ $ \dfrac{-12+12i}{-3-3i} = \dfrac{-12+12i}{-3-3i} \cdot \dfrac{{-3+3i}}{{-3+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-12+12i) \cdot (-3+3i)} {(-3-3i) \cdot (-3+3i)} = \dfrac{(-12+12i) \cdot (-3+3i)} {(-3)^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-12+12i) \cdot (-3+3i)} {(-3)^2 - (-3i)^2} = $ $ \dfrac{(-12+12i) \cdot (-3+3i)} {9 + 9} = $ $ \dfrac{(-12+12i) \cdot (-3+3i)} {18} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-12+12i}) \cdot ({-3+3i})} {18} = $ $ \dfrac{{-12} \cdot {(-3)} + {12} \cdot {(-3) i} + {-12} \cdot {3 i} + {12} \cdot {3 i^2}} {18} $ Evaluate each product of two numbers. $ \dfrac{36 - 36i - 36i + 36 i^2} {18} $ Finally, simplify the fraction. $ \dfrac{36 - 36i - 36i - 36} {18} = \dfrac{0 - 72i} {18} = -4i $